1. Equilibrium of Particles Free-body Diagram Equilibrium of Rigid Bodies
Lecture 6: Statics
Equilibrium of Particles
Free-body Diagram
Equilibrium of Rigid Bodies

2. Chapter 12 Static Equilibrium
Our whole built environment, from modern bridges to skyscrapers, has required architects and engineers to determine the forces and stresses within these structures. The object is to keep these structures static—that is, not in motion, especially not falling down.

3. Units of Chapter 12 The Conditions for Equilibrium
Solving Statics Problems

4. 12-1 The Conditions for Equilibrium
An object with forces acting on it, but with zero net force, is said to be in equilibrium.
Figure The book is in equilibrium; the net force on it is zero.
The first condition for equilibrium:

5. 12-1 The Conditions for Equilibrium
Example 12-1: Chandelier cord tension.p312
Calculate the tensions A and B in the two cords that are connected to the vertical cord supporting the 200-kg chandelier shown. Ignore the mass of the cords.
Solution: The forces at the point where the three cords join must add to zero. The vertical component of FA is the weight; the horizontal components of FA and FB cancel. Therefore, FA = 2260 N and FB = 1130 N.

6. 12-1 The Conditions for Equilibrium
The second condition of equilibrium is that there be no torque around any axis; the choice of axis is arbitrary.
Figure Although the net force on it is zero, the ruler will move (rotate). A pair of equal forces acting in opposite directions but at different points on an object (as shown here) is referred to as a couple.

7. Second Condition of Equilibrium
  = 0
= r  F
(r  F) = 0
 (r F sin ) = 0

8. Tips to carry out cross products
Put the tail of vector r & vector F at the same point.
Cross the four fingers from vector r to vector F through the smaller angle between the two vectors.
The thumb will give the direction of the cross product.
To be in equilibrium the magnitude of the cross products pointing in opposite directions must be equal.

9. 12-2 Solving Statics Problems
Choose one object at a time, and make a free-body diagram by showing all the forces on it and where they act.
Choose a coordinate system and the origin and resolve the components of the forces.
Write equilibrium equations for the forces.
Write the torque equilibrium equation. A clever choice of origin can simplify the problem enormously.
Solve.

10. 12-2 Solving Statics Problems
Example 12-3: Balancing a seesaw.
A board of mass M = 2.0 kg serves as a seesaw for two children. Child A has a mass of 30 kg and sits 2.5 m from the pivot point, P (his center of gravity is 2.5 m from the pivot). At what distance x from the pivot must child B, of mass 25 kg, place herself to balance the seesaw? Assume the board is uniform and centered over the pivot.
Figure (a) Two children on a seesaw. (b) Free-body diagram of the board.
Solution: Choose to take the axis of rotation to be the pivot point. We have two equations, the force equation and the torque equation. The force equation does not give us x, but the torque equation does; x = 3.0 m.

11. 12-2 Solving Statics Problems
If a force in your solution comes out negative (as A will here), it just means that it’s in the opposite direction from the one you chose. This is trivial to fix, so don’t worry about getting all the signs of the forces right before you start solving.
Figure A cantilever. The force vectors shown are hypothetical—one may even have a different direction.

12. 12-2 Solving Statics Problems
Example 12-5: Hinged beam and cable. p316
A uniform beam, 2.20 m long with mass m = 25.0 kg, is mounted by a small hinge on a wall. The beam is held in a horizontal position by a cable that makes an angle θ = 30.0°. The beam supports a sign of mass M = 28.0 kg suspended from its end. Determine the components of the force H that the (smooth) hinge exerts on the beam, and the tension FT in the supporting cable.
Solution: We have three unknowns, FT and the two components of FH. We know the angle of the cable, though, so we really need find only one component of FH. Taking the torque around the point where FT and mg act, and using the force equations, we find FHy = 123 N, FHx = 687 N, and FT = 793 N.

13. 12-2 Solving Statics Problems
Example 12-6: Ladder. p317
A 5.0-m-long ladder leans against a smooth wall at a point 4.0 m above a cement floor. The ladder is uniform and has mass m = 12.0 kg. Assuming the wall is frictionless (but the floor is not), determine the forces exerted on the ladder by the floor and by the wall.
Solution: We have three unknowns, the force exerted by the wall and the two forces exerted by the floor (the normal force and the static frictional force). The normal force equals the weight, 118 N (from the vertical force equation), and the static frictional force equals the force exerted by the wall; writing the torque equation around the point where the ladder touches the floor gives FW = 44 N.

14. Summary
An object at rest is in equilibrium; the study of such objects is called statics.
In order for an object to be in equilibrium, there must be no net force on it along any coordinate, and there must be no net torque around any axis.